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Which of these, if dissolved in 1.0 l of pure water, would produce a buffer solution? which of these, if dissolved in 1.0 l of p

stiks02 [169]

Answer:

(0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄)

Explanation:

A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or weak base and its conjugate acid.

Also, a buffer solution is a solution which resists changes in pH when acid or alkali is added to it.

(0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄) when dissolved in 1 L H₂O will produce a buffer because NaH₂PO₄ is considered as weak acid while Na₂HPO₄ is its conjugate base. 2.
(0.1 mol H₃O⁺ + 0.1 mol Cl⁻) is not a mixture of a weak acid and its conjugate base, or weak base and its conjugate acid.
(0.1 mol HCl + 0.1 mol NaoH) HCL is a strong acid and NaOH is a strong base so it will not form a buffer when dissolved in water.
(0.1 mol H₃O⁺ + 0.1 OH⁻) is not a mixture of a weak acid and its conjugate base, or weak base and its conjugate acid
(0.1 mol NaCl+ 0.1 mol KCl) NaCL and KCL are salts so it will not form a buffer when dissolved in water.

So the right choice is

(0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄)

7 0

3 years ago

¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.

irina [24]

Answer:

gfchjklnmbvh

Explanation:

4 0

3 years ago

Read 2 more answers

Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde

pochemuha

Solution :

We know that :

$\Delta T_f = k_f.m$ and $m=\frac{w_2}{m_2 \times w_1}$

Then, $\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$ ..................(1)

Where,

$w_1$ = amount of solvent (in kg)

$w_2$ = amount of solute (in kg)

$m_2$ = molar mass of solute (g/mole)

$m$ = molality of solution (mole/kg)

Given :

$\Delta T_f$ = $3.14\ ^\circ C$ , $k_f= 5.12\ ^\circ C/m$

$=5.12 \ ^\circ C/mole/kg$

$=5.12 \ ^\circ C \ kg/mole$

$w_1$ = 0.250 kg, $w_2$ = 24.3 g

Then putting this values in the equation is (1),

$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$

$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$

$m_2= 158.49$ g/mole

So, the molar mass of the unknown compound is 158.49 g/mole.

7 0

3 years ago

Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

dybincka [34]

Answer:

0.29

Explanation:

Step 1: Write the balanced equation at equilibrium

3 NO₂(g) + H₂O(l) ⇄ 2 HNO₃(aq) + NO(g)

Step 2: Calculate the molar concentration of the substances

We will not calculate the molarity of H₂O because pure liquids are not included in the equilibrium constant. We will use the following expression.

Molarity = mass of solute / molar mass of solute × liters of solution

Since the volume of the vessel is missing, we will assume it is 1 L to see the procedure.

[NO₂] = 22.5 / 46.01 g/mol × 1 L = 0.489 M

[HNO₃] = 15.5 g / 63.01 g/mol × 1 L = 0.246 M

[NO] = 16.6 g / 30.01 g/mol × 1 L = 0.553 M

Step 3: Calculate the value of the equilibrium constant Kc for this reaction

Kc = [HNO₃]² × [NO] / [NO₂]³

Kc = 0.246² × 0.553 / 0.489³

Kc = 0.29

7 0

3 years ago

A sample of gallium bromide, GaBr3, weighing 0.165 g was dissolved in water and treated with silver nitrate, AgNO3, resulting in

melamori03 [73]

Answer:

%Ga by mass in sample of $GaBr_{3}$ is 23.0%

Explanation:

Molar mass of AgBr = 187.77 g/mol

So, 0.299 g of AgBr = $\frac{0.299}{187.77}$ moles of AgBr = $\frac{0.299}{187.77}$ moles of Br = 0.00159 moles of Br

Br in AgBr comes from $GaBr_{3}$

So, there are 0.00159 moles of Br in 0.165 g of $GaBr_{3}$

Molar mass of Br = 79.904 g/mol

So, mass of Br in 0.165 g of $GaBr_{3}$ = $(0.00159\times 79.904)g$ = 0.127 g

So, mass of Ga in sample of $GaBr_{3}$ = (0.165-0.127) g = 0.038 g

So, %Ga by mass in sample of $GaBr_{3}$ = [(mass of Ga)/(mass of $GaBr_{3}$ )] $\times 100$ %

= $\frac{0.038}{0.165}\times 100$ %

= 23.0%

8 0

3 years ago